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A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5

1)+2):If there were originally 5 glazed doughnuts in the box, and one wasselected, then there are 4 glazed doughnuts in the box. Since theprobability of picking another glazed doughnut is 2/5, you know that 2/5 = 4 / num of d left, sothere must be a total of 10 doughnuts left in the box. So there were originally 5 glazed doughnuts out of 11 total.

Statement 1 just says that after removing 1 glazed doughnut the probability is 2/5. which is essentially 40%. Which means there could 4/10 glazed doughnuts making orig. prob = 5/10 = 1/2 or say 8/20 in which case the orig. prob = 9/21 or 3/7

Statement 1 just says that after removing 1 glazed doughnut the probability is 2/5. which is essentially 40%. Which means there could 4/10 glazed doughnuts making orig. prob = 5/10 = 1/2 or say 8/20 in which case the orig. prob = 9/21 or 3/7

Hence INSUFF

Statement 2 says there were 5 glazed doughnuts originallyNOT SUFF

Combining, let x be the total no. of doughnuts, then

2/5 x = 4 (since we know that there were 5 originally)

i.e. x = 20/2 = 10

Hence required prob. = 5/10 = 0.5

Correct the answer is C and your explanation matches except you forgot to add 1 to 10 making it P= 5/11.

But I still don't understand. If after removing 1st donut P=2/5, if we want to go back 1 step can't we just add 1 to both num and denom making it 3/6 =1/2?

A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

(2) There were originally 5 glazed doughnuts in the box.

Agree with C.

togather we know that there are 11 doughnuts altogather, out of which 5 are glazed...so the prob of getting a glazed doughnut at first draw = 5/11Therefore, the prob of getting a glazed doughnut at second draw = 4/10 = 2/5 _________________

A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

(2) There were originally 5 glazed doughnuts in the box.

Agree with C.

togather we know that there are 11 doughnuts altogather, out of which 5 are glazed...so the prob of getting a glazed doughnut at first draw = 5/11Therefore, the prob of getting a glazed doughnut at second draw = 4/10 = 2/5

I guess I over did the problem. If the first DN was put back and the probability of drawing the second DN is 2/5 then the glazed DN's count cannot be 5

A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

(2) There were originally 5 glazed doughnuts in the box.

(1)say there are x glazed donuts, and "n" total donutsfor the first glazed donut prob = x/nprob for second = x-1/n-1 = 2/5 But as this is ratio we can not conclude anything about x and n (it can be anythign 2/5, 10/25 etc)insuff

A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

(2) There were originally 5 glazed doughnuts in the box.

Agree with C.

togather we know that there are 11 doughnuts altogather, out of which 5 are glazed...so the prob of getting a glazed doughnut at first draw = 5/11Therefore, the prob of getting a glazed doughnut at second draw = 4/10 = 2/5

I guess I over did the problem. If the first DN was put back and the probability of drawing the second DN is 2/5 then the glazed DN's count cannot be 5

You are correct. The question is silent about the with or without replacement and therefore creats ambiguity to the takers. _________________

You are correct. The question is silent about the with or without replacement and therefore creats ambiguity to the takers.

IMHO, if the question does not specifically mention replacement => no replacement. I am not sure but it just occurred to me that I have read this somewhere......

Actually, the question would have speicified that the second probability is after the replacement, then there was no value in the question. But again.. I think it is responsibility of the question maker to avoid any ambiguity to us.

I think that first statement gives the probability of selecting second doughnut without putting it back. Thus, together with statement 2 it gives us the number of total doughnuts and the number of glazed doughnuts. _________________

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5.

This does NOT mean that the first chance is 3/6 = 1/2 because if there were, for example, 4 glazed and 6 other donuts after the first draw (4/10 = 2/5 chance), then before that there were 5 glazed and 6 others (5/11 chance). Insufficient.

(2) There were originally 5 glazed doughnuts in the box.

Obviously insufficient on its own since we need to know the total number BUT, with the first one it is sufficient:

If there were 5 glazed and one was drawn, there are now 4 glazed left. There's a 2/5 chance to draw another glazed so there are 4 glazed and 6 others (10 total). So before there were 5 glazed and 6 others (5/11 chance).

Combining these two statements, we have the value of 'g' from (2), which can be substituted in the equation obtained from (1) to find the value of 'n'.Thus, we can find out (g/n). SUFFICIENT. _________________

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5

1)+2):If there were originally 5 glazed doughnuts in the box, and one wasselected, then there are 4 glazed doughnuts in the box. Since theprobability of picking another glazed doughnut is 2/5, you know that 2/5 = 4 / num of d left, sothere must be a total of 10 doughnuts left in the box. So there were originally 5 glazed doughnuts out of 11 total.

A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5

1)+2):If there were originally 5 glazed doughnuts in the box, and one wasselected, then there are 4 glazed doughnuts in the box. Since theprobability of picking another glazed doughnut is 2/5, you know that 2/5 = 4 / num of d left, sothere must be a total of 10 doughnuts left in the box. So there were originally 5 glazed doughnuts out of 11 total.

Thank you kindly.

Experts,

Could you pls help to solve this DS.

Regards,Ammu

A box of doughnuts contains glazed, cream-filled, and jelly doughnuts. The box contains no other types of doughnuts. One doughnut is randomly selected from the box. What is the probability that the doughnut selected is glazed?

(1) If a glazed doughnut is selected, then the probability of withdrawing a second glazed doughnut is 2/5. We could have total of 6 doughnuts out of which 3 were glazed or 11 doughnuts out of which 5 were glazed or 16 doughnuts out of which 7 were glazed ... Not sufficient.

(2) There were originally 5 glazed doughnuts in the box. We don't know the total number doughnuts. Not sufficient.

(1)+(2) If there were originally 5 glazed doughnuts, then after selecting 1 of them there would be 4 left, so according to (1) 4/(total - 1) = 2/5 --> total = 11 --> P(glazed) = (glazed)/(total) = 5/11. Sufficient.